From: Loyer, Jeff W (firstname.lastname@example.org)
Date: Tue Nov 14 2000 - 10:28:57 PST
A question came up here that goes something like this:
If I'm going to buy a filter to slow my TDR pulse down to better mimic the
bus driver, should I slow it down to match the "slew-rate" or "rise-time"?
For instance, a drive signal going from 1.0V to 1.8V in 0.4nS has a
slew-rate of 2V/nS, and a rise-time of 0.4nS.
To mimic this driver's slew-rate, I would want my TDR (which has a 0.5V
voltage swing) to go from
-0.5V to 0V in 0.25nS. I would specify a 0.25nS rise-time filter.
To mimic this driver's rise-time, I would specify a 0.4nS rise-time filter.
Which is more important, the slew-rate or rise-time?
The first intuitive answer is that slew-rate is the important characteristic
- that a signal with higher slew-rate has more high-frequency components
than one with a lower slew-rate. Unfortunately, this implies that our
"rules of thumb" which talk about rise-time only are assuming a specific
voltage swing (and therefore a specific slew-rate for that rise-time), and
valid only for specific voltage swing levels.
The non-intuitive answer is that rise-time is the important characteristic,
regardless of slew-rate. This leaves the rules-of-thumb intact, but
violates our gut sense that a signal switching from 0 to 5V in 1nS is much
more "aggressive" (and sensitive to smaller discontinuities on a
transmission line, for instance) than one switching from 0 to 1V in 1nS.
My conclusion at this point is that the correct thing to do is to mimic the
rise-time, not the slew-rate.
Some simple simulations show that doubling the risetime cuts the ringback in
half, while doubling the slewrate has no effect.
Does anybody know of a good explanation (more than mere "gut feel" etc.) of
why the rise-time is the critical parameter, not slew-rate?
Note: at this point, I believe the major flaw in thinking slew-rate is
important is in the sentence "...a signal with higher slew-rate has more
high-frequency components than one with a lower slew-rate." I'm pretty sure
that, when you do the Fourier Transform of a pulse, the amplitude of the
pulse is merely a constant and comes out of the integration.
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